I think you reasoned, absolutely correctly, as follows. We have $x=2\pm\sqrt{1-y}^2$. Look at a slice at height $y$, of width "$dy$." The cross-section at height $y$ is a circle of radius $2+\sqrt{1-y^2}$, with a hole of radius $2-\sqrt{1-y^2}$.

The cross sectional area is therefore
$$\pi\left(2+\sqrt{1-y^2}\right)^2-\pi\left(2-\sqrt{1-y^2}\right)^2.$$
Simplify, by expanding the two squares. There is some nice cancellation, and we end up with $8\pi\sqrt{1-y^2}$. So the volume of our slice of thickness $dy$ is about $8\pi\sqrt{1-y^2}\,dy$. "Add up" (integrate) from $-1$ to $1$. We conclude that the volume of the torus is
$$8\pi\int_{-1}^1 \sqrt{1-y^2}\,dy.$$

It remains to evaluate the integral. There are two ways, one easy and the other harder.

Let's first do it the easy way. Think about $\int_{-1}^1 \sqrt{1-y^2}\,dy$, or, equivalently, about its good friend $\int_{-1}^1\sqrt{1-x^2}\,dx$.

The integral involving $x$ is exactly the integral you would write down if you wanted to find the area of the top half of the circle $x^2+y^2=1$. But the top half of this circle has area $\dfrac{\pi}{2}$. Multiply by $8\pi$, and we are finished!

But for completeness, let's do it the harder way. By symmetry, our integral is $2$ times $\int_0^1\sqrt{1-y^2}\,dy$. Make the substitution $y=\sin t$. Then
$$\int_0^1\sqrt{1-y^2}\,dy=\int_0^{\pi/2} \cos^2 t\,dt.$$
One relatively straightforward way to integrate $\cos^2 t$ is to use the double angle identity $\cos 2t =2\cos^2 t-1$ to rewrite $\cos^2 t$ as $\dfrac{\cos t +1}{2}$. Now the integration is not difficult. The definite integral of the $\cos 2t$ part is $0$.